Let $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$,and $\vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k}$. If the vector $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$,then $x$ equals:

$\vec{a}$ is a vector perpendicular to the plane containing non-zero vectors $\vec{b}$ and $\vec{c}$. If $\vec{a}, \vec{b}, \vec{c}$ are such that $|\vec{a}+\vec{b}+\vec{c}|=\sqrt{|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2}$, then $|(\vec{a} \times \vec{b}) \cdot \vec{c}|+|(\vec{a} \times \vec{b}) \times \vec{c}|=$

If $a \cdot b = b \cdot c = c \cdot a = 0$,then what is the value of the scalar triple product $[a b c]$?

Let $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}-\hat{k}$. If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=11$,$\vec{b} \cdot(\vec{a} \times \vec{c})=27$ and $\vec{b} \cdot \vec{c}=-\sqrt{3}|\vec{b}|$,then $|\vec{a} \times \vec{c}|^2$ is equal to

$(a + b) \cdot (b + c) \times (a + b + c) = $

If $a, b,$ and $c$ are unit coplanar vectors,then the scalar triple product $[a, b, c]$ is equal to: